Past exam of the representation theory of semisimple Lie algebras course of the University of Oxford / 2010 / 1

Put $e=E_{+}$ and $f=E_{-}$. The supplied identities giveIf no $f^{k}v$ vanishes, this is the first alternative. Otherwise choose the least $N\ge 1$ with $f^{N}v=0$. Applying $e$ gives $N(\chi -N+1)f^{N-1}v=0$. Since $f^{N-1}v\ne 0$ and the field has characteristic zero, $\chi =N-1$. ThusThe endpoint must include zero: the trivial one-dimensional module has $\chi =0$ and $fv=0$, so the printed claim of strict positivity has that counterexample. All nonzero vectors in the displayed chain have distinct $H$-weights, and hence are linearly independent.

For the intended module, index a basis by $v_k=a_{l-2k}$ for $0\le k\le l$, and usewith $v_{-1}=v_{l+1}=0$. This makes the basis endpoints distinct, identifies the lowering operator, and includes its missing raising action at $k=l$. These are the necessary indexing corrections to the printed display. The same definition for $l=0$ supplies the trivial module needed for classification.

Directly on $v_k$,These hold at the endpoints too, proving the three defining ${\mathfrak{sl}}_2$ relations. FinallyThus the Casimir element acts by that scalar on the whole module.

Let $V\ne 0$ be finite-dimensional. Choose an $H$-eigenvector and apply $e$ repeatedly until the last nonzero vector $v$ in that chain. Such a last vector exists because $e$ raises the $H$-eigenvalue by two and $H$ has only finitely many eigenvalues. Then $ev=0$ and $Hv=lv$ for some $l$.

The nonzero $f$-chain cannot continue indefinitely, since its weights are distinct. Part (a), with its zero endpoint included, gives $l\in {\mathbb{Z}}_{\ge 0}$ and the basis $v,fv,\dots ,f^{l}v$. The raising formula in part (a) shows its span is a submodule isomorphic to $A_l$.

To prove $A_l$ irreducible, a nonzero submodule stable under diagonal $H$ contains a weight vector: polynomial interpolation in the distinct eigenvalues projects onto individual representation weight lines. Repeated raising reaches $v_0$, and repeated lowering then reaches every $v_k$. Thus the submodule is all of $A_l$. Distinct $l$ give different dimensions $l+1$, so the modules are nonisomorphic. Every finite-dimensional irreducible module contains one of these nonzero simple submodules and hence equals it. ThereforeThe zero module has no nonzero simple submodule and is dealt with as the empty direct sum in part (d).

We first prove that every extension of two simple modules splits. ConsiderIf $n\ne m$, the central Casimir element has distinct scalars $n(n+2)$ and $m(m+2)$. On $V$ its minimal polynomial divides the product of the corresponding two distinct linear factors. Its eigenspaces are submodules, and the eigenspace for $m(m+2)$ maps isomorphically onto the quotient. It is the required complement.

Now suppose $n=m=l$. Use generalized weight spaces for $H$; their possible weights are $l,l-2,\dots ,-l$, because the characteristic polynomial on an extension is the product of those on its two factors. Lift the quotient's highest vector to $w$ in the generalized weight space of representation weight $l$. Then $ew=0$ and $f^{l+1}w=0$, since these would lie in generalized weight spaces of weights $l+2$ and $-l-2$, which are absent. Also $(H-l)w=a{v}_0$ for a highest vector $v_0$ in the submodule $A_l$. The commutator identity givesThus $a=0$. The vector $w$ is an actual highest-weight vector. Its lowering chain maps to the full basis of the quotient, so it generates an $A_l$ complementary to the original submodule. This proves splitting also when the two simple types coincide, including $l=0$.

The same splitting holds for a semisimple kernel and simple quotient by induction on the number of kernel summands. Write the kernel as $U_1\oplus U_2$ with $U_1$ simple. Modulo $U_2$, split the extension by the simple quotient; the inverse image of that chosen quotient copy is an extension with kernel $U_2$. Split it inductively, obtaining a submodule in $V$ meeting the entire original kernel trivially.

Finally induct on $\dim V$. A maximal proper submodule has smaller dimension and hence is semisimple by induction; the quotient is simple by maximality. The just-proved splitting completes the induction. Hencewith only finitely many nonzero multiplicities. This proof does not assume the complete-reducibility theorem it is establishing.